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PROFESSOR: Welcome back
to recitation.

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In this video segment, what I'd
like us to do is work on

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this following problem.

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Find d dx of the integral from
0 to x squared, cosine t dt.

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I'm going to give you a moment
to think about it and then

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I'll come back and show
you how I do it.

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OK, welcome back.

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Hopefully you were able to make
some headway on this.

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Let's look at the problem
and see how we

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would break it down.

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Well we know from the
fundamental theorem of

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calculus that you saw in the
lecture, that if up here

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instead of x squared we had an
x, then the problem would be

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easy to solve.

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We'd just use the fundamental
theorem of calculus, the

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answer would be cosine x.

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But of course, we don't have
an x, we have an x squared.

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That's why I gave you
this problem.

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And we need to figure out how
to solve this problem when

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there's a different function
up here besides just x.

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What we're going to use is we're
going to combine the

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fundamental theorem of calculus
and the chain rule.

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So, let's start off with how
we would do this if it were

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the integral from 0 to
x, as I mentioned.

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So we'll define capital F of x
to be equal to the integral

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from 0 to x cosine t dt.

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And then we know that f prime
of x is equal to cosine x.

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Now the problem is,
we don't just

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have this, as I mentioned.

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What we actually have--

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let me write this down--

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we have f of x squared.

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Right?

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That's what this-- sorry let
me highlight what I mean--

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this boxed thing is
f of x squared.

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So we took f of x and we
evaluated it at x squared.

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That's what we get in the box.

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And so if we want to find d dx
of f of x squared, it really

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is just the chain rule.

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We really want to think
of this as a

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composition of functions.

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The first, the outside function
is capital F and the

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inside function is x squared.

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So just in general, how do we
think about the chain rule?

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Well remember what we do--

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let me come back here
for a second--

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remember what we do is we take
the derivative of the outside

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function, we evaluate it at the
inside function and then

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we take the derivative of the
inside function and we

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multiply those two together.

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So all I have to do is figure
out, is what is

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the following thing?

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We know d dx, the quantity f of
x squared should be equal

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to F prime evaluated at
x squared times 2x.

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That's just what we said
earlier, right?

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It's the derivative of F
evaluated at x squared times

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the derivative of x squared.

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So now I just have to figure
out what this is.

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Well let's go back to the other
side and see what we

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wrote F prime was.

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If we come over here, we see F
prime at x is just cosine x.

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So F prime at x squared is
going to be this function

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evaluated at x squared.

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That's just cosine
of x squared.

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So we see over here, we just get
cosine x squared times 2x.

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And because I'm a mathematician,
I want to write

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the 2x in front before
I finish.

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Becaues otherwise
I get confused.

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So the answer here is just 2x
times cosine x squared.

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Now I want to point out really
what we did here.

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This is the answer to this
particular problem, but we can

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now solve problems in general
when I put any function up

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here, any function
of x up here.

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Right?

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Ultimately, all I did was I used
the fundamental theorem

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of calculus and the
chain rule.

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So any function I put
up here, I can do

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exactly the same process.

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I would define f of x to be this
type of thing, the way we

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would define it for the

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fundamental theorem of calculus.

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I would know what F
prime of x was.

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And then I would have to
evaluate F at a, at this

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function up here, whatever
I put up there.

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So in this case it
was x squared.

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I could have made it
natural log x.

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I could've made it some big
polynomial or something more

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complicated.

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Right?

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And once I do that, I just
follow this same process.

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Now, instead of the x
squared here I would

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have that other function.

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So I'd evaluate capital F at
whatever function that is

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times the derivative
of that function.

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It's exactly the same process.

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So I want to point out that
this is a bigger situation

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than I had before, or a bigger
situation than just this

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little problem.

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So, just so you understand
that.

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OK?

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So again, I just want to
say one more time.

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Now you know how to solve
problems where you have any

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other function of x up here
and you want to take the

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derivative of this kind of
expression of an integral with

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another function
of x up there.

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All right, I think
I'll stop there.

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